Question: $f(x, y) = y - \sin(x) - x^2$ What is the partial derivative of $f$ with respect to $x$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $1 - \cos(x) - 2x$ (Choice B) B $1$ (Choice C) C $-\cos(x) - 2x$ (Choice D) D $0$
Solution: Taking a partial derivative with respect to $x$ means treating $y$ like a constant, then taking a normal derivative. $\begin{aligned} \dfrac{\partial f}{\partial x} &= \dfrac{\partial}{\partial x} \left[ y - \sin({x}) - {x^2}\right] \\ \\ &= \dfrac{\partial}{\partial x} \left[ y \right] - \dfrac{\partial}{\partial x} \left[ \sin({x}) \right] - \dfrac{\partial}{\partial x} \left[ {x^2} \right] \\ \\ &= 0 - \cos({x}) - {2x} \end{aligned}$ In conclusion, $\dfrac{\partial f}{\partial x} = -\cos(x) - 2x $